Golden State Warriors forward David Lee has been named the NBA's Western Conference Player of the Week, according to 95.7 The Game's John Dickinson.
Lee helped the Warriors go 3-1 over the last seven days, and played a key role in major wins in Miami and Atlanta. During that stretch, he averaged 22.8 points on 60.6 percent from the field, 12.5 rebounds, 2.8 assists and 1.0 steals. He's had a double-double in seven consecutive games, which is the longest stretch in the NBA this season.
Lee beat out Blake Griffin of the Los Angeles Clippers, LaMarcus Aldridge of the Portland Trail Blazers, Kevin Durant of the Oklahoma City Thunder and Tony Parker of the San Antonio Spurs to claim the award. In the three wins, he averaged a point-rebound double-double.
Next on the schedule for Golden State are the New Orleans Hornets at home, where they will try to build off one of the most successful road trips in team history that saw the team go 6-1.